Question Number 105231 by mathmax by abdo last updated on 27/Jul/20 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$ Answered by 1549442205PVT last…
Question Number 39660 by abdo mathsup 649 cc last updated on 09/Jul/18 $${let}\:\:{S}_{{n}} \:\:=\:\int_{\mathrm{0}} ^{{n}} \:\:\:\:\:\frac{{x}\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\left({x}+\mathrm{1}\:−\left[{x}\right]\right)^{\mathrm{3}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{S}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} \\ $$…
Question Number 170725 by Sotoberry last updated on 29/May/22 Answered by FelipeLz last updated on 30/May/22 $${u}\:=\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow\:{du}\:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${dv}\:=\:\mathrm{1}{dx}\:\Rightarrow\:{v}\:=\:{x} \\ $$$$\int\mathrm{tan}^{−\mathrm{1}} \left({x}\right){dx} \\…
Question Number 170726 by Sotoberry last updated on 29/May/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170722 by Sotoberry last updated on 29/May/22 Answered by MikeH last updated on 29/May/22 $${u}\:=\mathrm{sin}\:\theta \\ $$$${du}\:=\:\mathrm{cos}\:\theta\:{d}\theta \\ $$$${dv}\:=\:{e}^{\theta} {d}\theta \\ $$$${v}\:=\:{e}^{\theta} \\…
Question Number 39633 by abdo mathsup 649 cc last updated on 09/Jul/18 $${find}\:{the}\:{value}\:{of}\: \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\pi} {ln}\left({x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cos}\theta\:+\mathrm{1}\right){d}\theta\:\:{with}\:{x}\:{fromR}. \\ $$ Commented by math khazana by…
Question Number 170663 by kndramaths last updated on 28/May/22 Commented by kaivan.ahmadi last updated on 28/May/22 $${a}. \\ $$$$\mathrm{0}\leqslant{x}\leqslant{a}\:\:,\:\:\mathrm{0}\leqslant{y}\leqslant\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\int\frac{{dx}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{{a}}{arctg}\left(\frac{{x}}{{a}}\right)…
Question Number 170610 by Sotoberry last updated on 27/May/22 Answered by aleks041103 last updated on 27/May/22 $${IBP} \\ $$$$\int{udv}={uv}−\int{vdu} \\ $$$${u}={x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$${v}={e}^{{x}} \\…
Question Number 170593 by 119065 last updated on 27/May/22 Commented by mkam last updated on 27/May/22 $$\int\:\left({x}+\mathrm{3}\:−\:\mathrm{3}\right)\:\left({x}+\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}=\int\:\left({x}+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{3}\left({x}+\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$ \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{5}}\:\left({x}+\mathrm{3}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:−\:\mathrm{2}\:\left({x}+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}}…
Question Number 170588 by bagjagugum123 last updated on 27/May/22 Terms of Service Privacy Policy Contact: info@tinkutara.com